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primary 5 | Maths
| Whole Numbers
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Date Posted: 3 years ago
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Given that there are 7 mangoes more than watermelon bought, the least number of mangoes bought will be 8 and 1 watermelon.
However, the cost of this does not tally and the numbers, so by trial and error. Increase the number of mangoes by 1 each time..maintaining the difference in number (7) each time
Until, 11 mangoes and 4 watermelons.
cost of 11 mangoes = 11 x $2 =$22
Cost of 4 watermelons = 4x $5 = $20
$66 -$42 = $24
Given cost of 7 apples = $4,
$24/4 = 6 sets of 7 apples = 6x7 = 42 apples
Check: Total number of fruits = 42 + 11+ 4 = 57
However, the cost of this does not tally and the numbers, so by trial and error. Increase the number of mangoes by 1 each time..maintaining the difference in number (7) each time
Until, 11 mangoes and 4 watermelons.
cost of 11 mangoes = 11 x $2 =$22
Cost of 4 watermelons = 4x $5 = $20
$66 -$42 = $24
Given cost of 7 apples = $4,
$24/4 = 6 sets of 7 apples = 6x7 = 42 apples
Check: Total number of fruits = 42 + 11+ 4 = 57
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